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A box with a mass of 10 $\mathrm{kg}$ is placed on an inclined plane that makes a $60^{\circ}$ angle with the horizontal. The coefficient of static fraction $\left(\mu_{\mathrm{s}}\right)$ between the box and the inclined plane is $0.2 .$ What force must be applied on the box in order to prevent the box from sliding down the inclined plane?(A) 32.7 $\mathrm{N}$(B) 48.3 $\mathrm{N}$(C) 56.2 $\mathrm{N}$(D) 76.6 $\mathrm{N}$

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Chapter 14

Practice Test 4

Section 1

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we have given a box of mask 10 kg is placed on a line plane. This angle is 60 degree, so we will do it. First of all, a three boy diagrams the M G force in downward direction Normal force in this direction and here mg sine theta mg sine 60 will be there and in this direction mg cost 60 will be there and in the right direction Friction force will be there and we know that there is no motion. That means networks on this block should be called to zero, so normal force should be equal to mg cost 60 right and mg signed 60 plus friction force should be equal to Yeah I suppose that force is equal to F So we will write equation in horizontal direction We will write net force should be equal to zero along the in plane So free f plus friction for should be equal to I am designed 60 So this f will be equal to mg sine 60 minus friction force which is new is mg course 63 to 60 degree angle So now we will put the values understand gs sign 60 minus us which is 0.2 into cost 60. Now we will put the values. If G is equal to 10 m per second square, then the value of force is equal to 76 point six Newton. So option devil, we were correct. Option. Thank you.

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